Yes they do. In order to do this you must make the fraction 70/2 and 105/3. If you divided each one of those you get 35 for both which means it's a proportion.
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
_____
My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
Since each term contains x, x= 0 is one answer Also as x^2 is a factor it has duplicity 2.
Factoring:-
x^2 (x^2 - 4x + 3) = 0
x^2( x - 3)(x - 1) = 0
so x = 3 and x = 1 are also zeros
Answer:- x = 0 (duplicity 2), x = 1 and x = 3.
Answer:
49%
Step-by-step explanation:
7 percent of 500=35%
